ALL 8 CHAPTERS COVERED
SOLUTION MANUAL
, Solutions for Chapter 1
Solutions for exercises in section 1. 2
1.2.1. (1, 0, 0)
1.2.2. (1, 2, 3)
1.2.3. (1, 0, −1)
1.2.4. (−1/2, 1/2, 0, 1)
2 −4 3
1.2.5. 4 −7 4
5 −8 4
1.2.6. Every row operation is reversible. In particular the “inverse” of any row operation
is again a row operation of the same type.
π
1.2.7. 2 , π, 0
1.2.8. The third equation in the triangularized form is 0x3 = 1, which is impossible
to solve.
1.2.9. The third equation in the triangularized form is 0x3 = 0, and all numbers are
solutions. This means that you can start the back substitution with any value
whatsoever and consequently produce infinitely many solutions for the system.
1.2.10. α = −3, β = 11 3
2 , and γ = − 2
1.2.11. (a) If xi = the number initially in chamber #i, then
.4x1 + 0x2 + 0x3 + .2x4 = 12
0x1 + .4x2 + .3x3 + .2x4 = 25
0x1 + .3x2 + .4x3 + .2x4 = 26
.6x1 + .3x2 + .3x3 + .4x4 = 37
and the solution is x1 = 10, x2 = 20, x3 = 30, and x4 = 40.
(b) 16, 22, 22, 40
1.2.12. To interchange rows i and j, perform the following sequence of Type II and
Type III operations.
Rj ← R j + Ri (replace row j by the sum of row j and i)
Ri ← Ri − Rj (replace row i by the difference of row i and j)
Rj ← R j + Ri (replace row j by the sum of row j and i)
Ri ← −Ri (replace row i by its negative)
1.2.13. (a) This has the effect of interchanging the order of the unknowns— xj and
xk are permuted. (b) The solution to the new system is the same as the
,2 Solutions
solution to the old system except that the solution for the j th unknown of the
new system is x̂j = α1 xj . This has the effect of “changing the units” of the j th
unknown. (c) The solution to the new system is the same as the solution for
the old system except that the solution for the k th unknown in the new system
is x̂k = xk − αxj .
1
1.2.14. hij = i+j−1
x1 y1
x2 y2
1.2.16. If x =
... and y = .. are two different solutions, then
.
xm ym
x1 +y1
2
x2 +y2
x+y
2
z= = ..
2 .
xm +ym
2
is a third solution different from both x and y.
Solutions for exercises in section 1. 3
1.3.1. (1, 0, −1)
(2, −1, 0, 0)
1.3.2.
1 1 1
1.3.3. 1 2 2
1 2 3
Solutions for exercises in section 1. 4
yk+1 − yk−1 yk−1 − 2yk + yk+1
1.4.2. Use y ′ (tk ) = yk′ ≈ and y ′′ (tk ) = yk′′ ≈ to write
2h h2
2yk−1 − 4yk + 2yk+1 hyk+1 − hyk−1
f (tk ) = fk = yk′′ −yk′ ≈ − , k = 1, 2, . . . , n,
2h2 2h2
with y0 = yn+1 = 0. These discrete approximations form the tridiagonal system
−4 2−h y1 f1
2 + h −4 2−h y2 f2
.. .. .. .. ..
. . . = 2h2 .
. .
2+h −4 2 − h
yn−1 fn−1
2+h −4 yn fn
, Solutions 3
Solutions for exercises in section 1. 5
1 −1
1.5.1. (a) (0, −1) (c) (1, −1) (e) 1.001 , 1.001
2
1.5.2. (a) (0, 1) (b) (2, 1) (c) (2, 1) (d) 1.0001 , 1.0003
1.0001
1.5.3. Without
PP: (1.01, 1.03) WithPP: (1,1) Exact: (1, 1)
1 .500 .333 .333 1 .500 .333 .333
1.5.4. (a) .500 .333 .250 .333 −→ 0 .083 .083 .166
.333 .250 .200 .200 0 .083 .089 .089
1 .500 .333 .333
−→ 0 .083 .083 .166 z = −.077/.006 = −12.8,
0 0 .006 −.077
y = (.166 − .083z)/.083 = 14.8, x = .333 − (.5y + .333z) = −2.81
1 .500 .333 .333 1 .500 .333 .333
(b) .500 .333 .250 .333 −→ 1 .666 .500 .666
.333 .250 .200 .200 1 .751 .601 .601
1 .500 .333 .333 1 .500 .333 .333
−→ 0 .166 .167 .333 −→ 0 .251 .268 .268
0 .251 .268 .268 0 .166 .167 .333
1 .500 .333 .333
−→ 0 .251 .268 .268 z = −.156/.01 = −15.6,
0 0 −.01 .156
y = (.268 − .268z)/.251 = 17.7, x = .333 − (.5y + .333z) = −3.33
1 .500 .333 .333 1 .500 .333 .333
(c) .500 .333 .250 .333 −→ 1 .666 .500 .666
.333 .250 .200 .200 1 .751 .601 .601
1 .500 .333 .333 1 .500 .333 .333
−→ 0 .166 .167 .333 −→ 0 .994 1 1.99
0 .251 .268 .268 0 .937 1 1
1 .500 .333 .333
−→ 0 .994 1 1.99 z = −.88/.057 = −15.4,
0 0 .057 −.880
y = (1.99 − z)/.994 = 17.5, x = .333 − (.5y + .333z) = −3.29
(d) x = −3, y = 16, z = −14
1.5.5. (a)
.0055x + .095y + 960z = 5000
.0011x + . 01y + 112z = 600
.0093x + .025y + 560z = 3000
SOLUTION MANUAL
, Solutions for Chapter 1
Solutions for exercises in section 1. 2
1.2.1. (1, 0, 0)
1.2.2. (1, 2, 3)
1.2.3. (1, 0, −1)
1.2.4. (−1/2, 1/2, 0, 1)
2 −4 3
1.2.5. 4 −7 4
5 −8 4
1.2.6. Every row operation is reversible. In particular the “inverse” of any row operation
is again a row operation of the same type.
π
1.2.7. 2 , π, 0
1.2.8. The third equation in the triangularized form is 0x3 = 1, which is impossible
to solve.
1.2.9. The third equation in the triangularized form is 0x3 = 0, and all numbers are
solutions. This means that you can start the back substitution with any value
whatsoever and consequently produce infinitely many solutions for the system.
1.2.10. α = −3, β = 11 3
2 , and γ = − 2
1.2.11. (a) If xi = the number initially in chamber #i, then
.4x1 + 0x2 + 0x3 + .2x4 = 12
0x1 + .4x2 + .3x3 + .2x4 = 25
0x1 + .3x2 + .4x3 + .2x4 = 26
.6x1 + .3x2 + .3x3 + .4x4 = 37
and the solution is x1 = 10, x2 = 20, x3 = 30, and x4 = 40.
(b) 16, 22, 22, 40
1.2.12. To interchange rows i and j, perform the following sequence of Type II and
Type III operations.
Rj ← R j + Ri (replace row j by the sum of row j and i)
Ri ← Ri − Rj (replace row i by the difference of row i and j)
Rj ← R j + Ri (replace row j by the sum of row j and i)
Ri ← −Ri (replace row i by its negative)
1.2.13. (a) This has the effect of interchanging the order of the unknowns— xj and
xk are permuted. (b) The solution to the new system is the same as the
,2 Solutions
solution to the old system except that the solution for the j th unknown of the
new system is x̂j = α1 xj . This has the effect of “changing the units” of the j th
unknown. (c) The solution to the new system is the same as the solution for
the old system except that the solution for the k th unknown in the new system
is x̂k = xk − αxj .
1
1.2.14. hij = i+j−1
x1 y1
x2 y2
1.2.16. If x =
... and y = .. are two different solutions, then
.
xm ym
x1 +y1
2
x2 +y2
x+y
2
z= = ..
2 .
xm +ym
2
is a third solution different from both x and y.
Solutions for exercises in section 1. 3
1.3.1. (1, 0, −1)
(2, −1, 0, 0)
1.3.2.
1 1 1
1.3.3. 1 2 2
1 2 3
Solutions for exercises in section 1. 4
yk+1 − yk−1 yk−1 − 2yk + yk+1
1.4.2. Use y ′ (tk ) = yk′ ≈ and y ′′ (tk ) = yk′′ ≈ to write
2h h2
2yk−1 − 4yk + 2yk+1 hyk+1 − hyk−1
f (tk ) = fk = yk′′ −yk′ ≈ − , k = 1, 2, . . . , n,
2h2 2h2
with y0 = yn+1 = 0. These discrete approximations form the tridiagonal system
−4 2−h y1 f1
2 + h −4 2−h y2 f2
.. .. .. .. ..
. . . = 2h2 .
. .
2+h −4 2 − h
yn−1 fn−1
2+h −4 yn fn
, Solutions 3
Solutions for exercises in section 1. 5
1 −1
1.5.1. (a) (0, −1) (c) (1, −1) (e) 1.001 , 1.001
2
1.5.2. (a) (0, 1) (b) (2, 1) (c) (2, 1) (d) 1.0001 , 1.0003
1.0001
1.5.3. Without
PP: (1.01, 1.03) WithPP: (1,1) Exact: (1, 1)
1 .500 .333 .333 1 .500 .333 .333
1.5.4. (a) .500 .333 .250 .333 −→ 0 .083 .083 .166
.333 .250 .200 .200 0 .083 .089 .089
1 .500 .333 .333
−→ 0 .083 .083 .166 z = −.077/.006 = −12.8,
0 0 .006 −.077
y = (.166 − .083z)/.083 = 14.8, x = .333 − (.5y + .333z) = −2.81
1 .500 .333 .333 1 .500 .333 .333
(b) .500 .333 .250 .333 −→ 1 .666 .500 .666
.333 .250 .200 .200 1 .751 .601 .601
1 .500 .333 .333 1 .500 .333 .333
−→ 0 .166 .167 .333 −→ 0 .251 .268 .268
0 .251 .268 .268 0 .166 .167 .333
1 .500 .333 .333
−→ 0 .251 .268 .268 z = −.156/.01 = −15.6,
0 0 −.01 .156
y = (.268 − .268z)/.251 = 17.7, x = .333 − (.5y + .333z) = −3.33
1 .500 .333 .333 1 .500 .333 .333
(c) .500 .333 .250 .333 −→ 1 .666 .500 .666
.333 .250 .200 .200 1 .751 .601 .601
1 .500 .333 .333 1 .500 .333 .333
−→ 0 .166 .167 .333 −→ 0 .994 1 1.99
0 .251 .268 .268 0 .937 1 1
1 .500 .333 .333
−→ 0 .994 1 1.99 z = −.88/.057 = −15.4,
0 0 .057 −.880
y = (1.99 − z)/.994 = 17.5, x = .333 − (.5y + .333z) = −3.29
(d) x = −3, y = 16, z = −14
1.5.5. (a)
.0055x + .095y + 960z = 5000
.0011x + . 01y + 112z = 600
.0093x + .025y + 560z = 3000
